Question: What is the average value of $x^3-9x$ on the interval $-1\leq x \leq 3$ ?
Explanation: In general, this is the average value of function $f$ over the interval $[a,b]$ : $\dfrac{\int_a^b f(x)\,dx}{b-a}$ In our case, ${f(x)=x^3-9x}$, ${a=-1}$ and ${b=3}$ : $\begin{aligned} \dfrac{\int_{ a}^{ b} {f(x)}\,dx}{ b- a}&=\dfrac{\int_{{-1}}^{ {3}} ({x^3-9x})\,dx}{{3}-{(-1)}} \\\\ &=\dfrac{\left[\dfrac{x^4}{4}-\dfrac{9x^2}{2}\right]_{-1}^{3}}{4} \\\\ &=\dfrac{-20.25-(-4.25)}{4} \\\\ &=-4 \end{aligned}$ In conclusion, the average value of $x^3-9x$ on the interval $-1\leq x \leq 3$ is $-4$.